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Inga [223]
3 years ago
15

A spin balancer rotates the wheel of a car at 480 revolutions per minute. if the diameter of the wheel is 26 inches, what road s

peed is being tested? express your answer in miles per hour. at how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?
Physics
1 answer:
Mariana [72]3 years ago
3 0
Part A:
To answer this item, we need to convert the given revolutions per minute in speed expressed as miles per hour. In order to do so, we need to use the dimensional analysis and the appropriate conversion factor.
 
We calculate the circumference that is covered by the wheel.
     C = πD
where C is the circumference and D is the diameter.

Substitute the known values,
       C = π(26 in) = 81.68 inches

The speed tested is calculated below.
      s = (81.68 inches/rev)(480 rev/min)(1 mil/63360 inches)(60 min/1 hr)
            s = 37.127 mil/h

<em>ANSWER: 37.127 mil/h</em>

Part B:
Similar to what was done in Part A, the setting of the balancer is calculated below,
       setting = (80 mil/h)(1h/60 min)(63360 in/1 mil)(1 rev/81.68 in)
         setting = 1034.23 rev/min

<em>ANSWER: 1034.23 rev/min</em>
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Answer:

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Explanation:

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As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

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\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

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4.3 x 9.46 x 10¹⁵ = (14000) t

t = 2.91 x 10¹² sec

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